Work, Heat, and Internal Energy in Isobaric & Isochoric Processes

Hello! Welcome to the first lesson of our module on Thermodynamic Processes.

In the last module, we explored the First Law of Thermodynamics, culminating in a deep dive into irreversible processes like free expansion. We saw that for an adiabatic free expansion of an ideal gas, the initial and final temperatures are the same ( $Δ T = 0$ ), even though the process is not isothermal. This distinction between a process's overall result ( $Δ T = 0$ ) and its path (isothermal means constant temperature throughout) is critical.

Today, we begin our systematic analysis of the four fundamental quasi-static (reversible) processes that form the building blocks of thermodynamics. This lesson focuses on the two simplest cases. Your goal is to be able to calculate the work done (W), heat transferred (Q), and change in internal energy ( $Δ U$ ) for isochoric and isobaric processes.

We will pay special attention to the sign conventions for work and heat, a topic you've highlighted as important for avoiding errors in JEE Advanced problems.

1. Isochoric Process (Constant Volume)

An isochoric process is defined as a thermodynamic process that occurs at a constant volume. Imagine a gas sealed in a rigid, strong container. If you heat it, its pressure and temperature will rise, but its volume cannot change.

a) Work Done ( $W$ )

The definition of work done by a gas is given by the integral $W = \int P d V$ .

In an isochoric process, the volume is constant, which means the change in volume $d V$ is zero. Consequently, the work done is always zero.

$W_{i soc h or i c} = \int_{V_{i}}^{V_{f}} P d V = 0 (since d V = 0)$

On a P-V diagram, an isochoric process is represented by a vertical line. The work done is the area under the P-V curve, which is clearly zero for a vertical line.

Isochoric Process Summary — This P-V diagram shows a vertical line, characteristic of an isochoric process. The key equations governing this process are summarized, highlighting that work done (W) is zero.

b) Internal Energy ( $Δ U$ ) and Heat (Q)

Let's apply the First Law of Thermodynamics (FLOT) using the physics convention we've been working with:
$Δ U = Q - W$

Since $W = 0$ for an isochoric process, the first law simplifies beautifully:
$Δ U = Q_{i soc h or i c}$

This means that any heat added to the system directly increases its internal energy, and any heat removed directly decreases it.

Now, how do we calculate this change? For an ideal gas, the change in internal energy is always given by:
$Δ U = n C_{v} Δ T$
where $n$ is the number of moles and $C_{v}$ is the molar specific heat at constant volume. It is crucial to remember that this formula for $Δ U$ is valid for any process involving an ideal gas, not just isochoric ones. Its name comes from the fact that it also represents the heat exchanged in a constant volume process.

Combining these gives us the expression for heat transferred:
$Q_{i soc h or i c} = n C_{v} Δ T$

To see these concepts applied in a problem-solving context, let's watch a video.

Isochoric Process Thermodynamics - Work, Heat & Internal Energy, PV Diagrams

The following video from The Organic Chemistry Tutor provides a clear, step-by-step walkthrough of the calculations for an isochoric process. It reinforces why work is zero and how to calculate temperature change, internal energy, and heat.

Watch the video from the beginning until 05:54. Pay close attention to: The reasoning for why work done is zero (00:00 - 01:24). How the ideal gas law is used to find the temperature change (01:24 - 04:01). The calculation of internal energy change using \Delta U = nC_v \Delta T (04:01 - 05:00). How the First Law simplifies to find the heat transferred, Q = \Delta U (05:00 - 05:54).

Summary for an Isochoric Process (Ideal Gas):

Condition: $Δ V = 0$
Work Done: $W = 0$
Internal Energy Change: $Δ U = n C_{v} Δ T$ (Always true for an ideal gas)
Heat Transferred: $Q = Δ U = n C_{v} Δ T$

2. Isobaric Process (Constant Pressure)

An isobaric process occurs at constant pressure. A common example is heating a gas in a cylinder with a freely moving piston under a constant external atmospheric pressure. As the gas is heated, it expands to keep its pressure equal to the external pressure.

a) Work Done ( $W$ )

We again start with the definition $W = \int P d V$ . Since pressure $P$ is constant, we can take it out of the integral:

$W_{i so ba r i c} = P \int_{V_{i}}^{V_{f}} d V = P (V_{f} - V_{i}) = P Δ V$

On a P-V diagram, this corresponds to the rectangular area under the horizontal line representing the process.

Given your background, you'll appreciate an alternative form derived from the ideal gas law, $P V = n R T$ . For a process at constant pressure:
$P Δ V = P (V_{f} - V_{i}) = P V_{f} - P V_{i} = n R T_{f} - n R T_{i} = n R Δ T$
Therefore, the work done can also be calculated as:
$W_{i so ba r i c} = n R Δ T$
This is a very useful formula in problems where temperature change is given instead of volume change.

b) Internal Energy ( $Δ U$ ) and Heat (Q)

As emphasized before, for an ideal gas, the change in internal energy depends only on the change in temperature, regardless of the process path:
$Δ U = n C_{v} Δ T$

Now, let's find the heat transferred using FLOT:
$Q = Δ U + W$
Substituting our expressions for $Δ U$ and $W$ :
$Q_{i so ba r i c} = n C_{v} Δ T + P Δ V$

This is often written in a more compact form using the molar heat capacity at constant pressure, $C_{p}$ :
$Q_{i so ba r i c} = n C_{p} Δ T$

By comparing the two expressions for $Q_{i so ba r i c}$ , we can derive a fundamental relationship.
$n C_{p} Δ T = n C_{v} Δ T + P Δ V$
Using $P Δ V = n R Δ T$ :
$n C_{p} Δ T = n C_{v} Δ T + n R Δ T$
Dividing by $n Δ T$ gives Mayer's Relation:
$C_{p} = C_{v} + R$
This relation beautifully captures the fact that at constant pressure, more heat is required to raise the temperature by the same amount than at constant volume, because some of the energy is used to do expansion work.

The following resource from Allen provides a good summary of these formulas.

Thermodynamics: Definition, Process, Equations & Laws - Allen

This webpage provides a concise summary of the formulas for different thermodynamic processes, which is useful for quick revision. I'm pointing you to a specific solved example that applies these concepts directly.

First, find the section titled 'Sample Questions On Thermodynamics'. Read question Q-2 and its solution. This is a classic isobaric process problem where you are given heat (\Delta Q) and need to find the change in internal energy (\Delta U) by first calculating the work done (P\Delta V). Then, look at section 12, 'Different Thermodynamic Process', and review the summary for the Isobaric Process.

This next video demonstrates how to tackle a more detailed isobaric problem from start to finish.

Isobaric Process Thermodynamics - Work & Heat Energy, Molar Heat Capacity, & Internal Energy

This video from The Organic Chemistry Tutor is a comprehensive guide to solving problems on isobaric processes. It covers all three calculations: W, Q, and \Delta U.

Please watch the following segments: Work Calculation (00:00 - 01:36): See the direct application of W = P\Delta V. Heat Calculation (08:21 - 11:33): Focus on how Q = nC_p \Delta T is used, and how to determine the value of C_p for a monoatomic gas. Internal Energy Calculation (11:33 - 13:20): Note how \Delta U is calculated in two ways: first using \Delta U = nC_v \Delta T and then verified using \Delta U = Q - W. This reinforces the consistency of the First Law.

Summary for an Isobaric Process (Ideal Gas):

Condition: $Δ P = 0$
Work Done: $W = P Δ V = n R Δ T$
Internal Energy Change: $Δ U = n C_{v} Δ T$
Heat Transferred: $Q = n C_{p} Δ T$

3. A Deep Dive into Sign Conventions

This is a major source of confusion and errors, and mastering it is key. The convention depends on the context, typically physics/engineering vs. chemistry.

Convention	First Law of Thermodynamics	Work (W)	Heat (Q)
Physics/Engineering	$Δ U = Q - W$	$W > 0$ : Work done by the system (expansion) $W < 0$ : Work done on the system (compression)	$Q > 0$ : Heat added to system $Q < 0$ : Heat removed from system
Chemistry (IUPAC)	$Δ U = Q + W$	$W > 0$ : Work done on the system (compression) $W < 0$ : Work done by the system (expansion)	$Q > 0$ : Heat added to system $Q < 0$ : Heat removed from system

Why the difference?

Physics/Engineering: The focus is often on engines, where the goal is to get useful work out of the system. So, work done by the system is considered positive.
Chemistry: The focus is on the energy changes of the molecules (the system). When the system does work, it expends its energy, so that change is considered negative from the system's point of view.

The following resource explains the IUPAC (chemistry) convention, which will help you solidify the contrast.

Chemical Thermodynamics Chapter for JEE Main Chemistry - Vedantu

To understand the alternative perspective, please review this FAQ section from a JEE resource by Vedantu. It clearly lays out the IUPAC sign convention used in chemistry.

Scroll down to the 'Frequently Asked Questions' section. Read the answers to questions 3 and 4. Focus on how 'w' is defined for an isobaric process (w = -P_{ext}\Delta V) and the explicit rules for the signs of 'w' and 'q'.

Your Strategy for JEE Problems:

Identify the Convention: Most physics problems will use the $Δ U = Q - W$ convention. Be alert for chemistry-based thermodynamic questions that might use $Δ U = Q + W$ .
Focus on the Words: The most reliable method is to ignore the sign temporarily and calculate the magnitude of the work, e.g., $∣ W ∣ = ∣ P Δ V ∣$ . Then, read the question carefully.
- If it asks for "work done by the gas" during expansion, the answer is positive.
- If it asks for "work done on the gas" during expansion, the answer is negative.
- The opposite is true for compression.
Be Consistent: Once you've chosen a convention for $W$ , use the corresponding version of the First Law ( $Δ U = Q - W$ or $Δ U = Q + W$ ) throughout the entire problem.

Test your understanding!

A diatomic ideal gas expands isobarically, doing 200 J of work on its surroundings. Calculate the heat supplied to the gas and the change in its internal energy.
(For a diatomic gas, $C_{v} = \frac{5}{2} R$ and $C_{p} = \frac{7}{2} R$ ).

Show answer

Let's use the physics convention ( $Δ U = Q - W$ ).

Work (W): The gas does work on its surroundings, so work done by the system is positive. $W = + 200$ J.
Relate W and $Δ U$ : We know $W = n R Δ T$ and $Δ U = n C_{v} Δ T$ . We can find a ratio:
$\frac{Δ U}{W} = \frac{n C _{v} Δ T}{n R Δ T} = \frac{C _{v}}{R}$
For a diatomic gas, $C_{v} = \frac{5}{2} R$ , so $\frac{C _{v}}{R} = \frac{5}{2}$ .
$Δ U = W \times \frac{C _{v}}{R} = 200 J \times \frac{5}{2} = 500 J$
Heat (Q): Using the First Law:
$Q = Δ U + W = 500 J + 200 J = 700 J$
Alternatively, using $Q = n C_{p} Δ T$ , we can find the ratio $\frac{Q}{W} = \frac{C _{p}}{R} = \frac{7/2 R}{R} = \frac{7}{2}$ .
$Q = W \times \frac{C _{p}}{R} = 200 J \times \frac{7}{2} = 700 J$
The results are consistent. Heat supplied is 700 J, and the change in internal energy is 500 J.

Conclusion

In this lesson, we established the core equations for analyzing two fundamental thermodynamic processes.

Key Takeaways:

Isochoric Process (V=const): No work is done ( $W = 0$ ). All heat exchanged contributes to internal energy ( $Q = Δ U = n C_{v} Δ T$ ).
Isobaric Process (P=const): Work is done ( $W = P Δ V = n R Δ T$ ). Heat supplied is used for both increasing internal energy and doing work ( $Q = n C_{p} Δ T = Δ U + W$ ).
Internal Energy of an Ideal Gas: The change $Δ U$ depends only on the temperature change ( $Δ U = n C_{v} Δ T$ ) for any process.
Sign Conventions: The definition of positive/negative work differs between physics and chemistry. Always be clear which convention you are using and what the question is asking for ("work done by" vs. "work done on").

Preview of the Next Lesson:
We will continue our journey by analyzing the remaining two key processes: isothermal (constant temperature) and adiabatic (no heat exchange). These processes involve non-linear curves on a P-V diagram and require calculus to derive the work done. We will also compare their P-V slopes, a classic topic for advanced exams.

Recommended resources

1. Isochoric Process (Constant Volume)

a) Work Done (W)

b) Internal Energy (ΔU) and Heat (Q)

2. Isobaric Process (Constant Pressure)

a) Work Done (W)

b) Internal Energy (ΔU) and Heat (Q)

3. A Deep Dive into Sign Conventions

Conclusion

a) Work Done ( $W$ )

b) Internal Energy ( $Δ U$ ) and Heat (Q)

a) Work Done ( $W$ )

b) Internal Energy ( $Δ U$ ) and Heat (Q)